`lim_(xrarroo)([x]+[2x]+[3x]+….+[nx])/(n^(2))`, where `[*]` denotes greatest integer function, is

Options

A. `(x)/(3)`
B. `(x)/(6)`
C. `(x)/(2)`
D. does not exist

Answer

To solve the limit problem

\[ \lim_{x \to \infty} \frac{[x] + [2x] + [3x] + \ldots + [nx]}{n^2} \]

where \([*]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Understanding the Greatest Integer Function
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). For large \(x\), we can express \([kx]\) as:
\[ [kx] = kx - \{kx\} \]
where \(\{kx\}\) is the fractional part of \(kx\).

### Step 2: Rewrite the Sum
We can rewrite the sum in the limit:
\[ [x] + [2x] + [3x] + \ldots + [nx] = (x + 2x + 3x + \ldots + nx) - (\{x\} + \{2x\} + \{3x\} + \ldots + \{nx\}) \]
The first part simplifies to:
\[ x(1 + 2 + 3 + \ldots + n) = x \cdot \frac{n(n + 1)}{2} \]

### Step 3: Analyze the Fractional Part
The second part, \(\{x\} + \{2x\} + \{3x\} + \ldots + \{nx\}\), is bounded. Specifically, since each \(\{kx\}\) is between 0 and 1, the sum \(\{x\} + \{2x\} + \ldots + \{nx\}\) is at most \(n\):
\[ 0 \leq \{x\} + \{2x\} + \ldots + \{nx\} < n \]

### Step 4: Combine the Results
Thus, we can express the limit as:
\[ \lim_{x \to \infty} \frac{\left( x \cdot \frac{n(n + 1)}{2} \right) - (\{x\} + \{2x\} + \ldots + \{nx\})}{n^2} \]
This simplifies to:
\[ \lim_{x \to \infty} \left( \frac{xn(n + 1)}{2n^2} - \frac{\{x\} + \{2x\} + \ldots + \{nx\}}{n^2} \right) \]

### Step 5: Evaluate the Limit
As \(x\) approaches infinity, the term \(\frac{xn(n + 1)}{2n^2}\) dominates:
\[ \frac{xn(n + 1)}{2n^2} = \frac{x(n + 1)}{2n} \]
The second term, \(\frac{\{x\} + \{2x\} + \ldots + \{nx\}}{n^2}\), approaches 0 since it is bounded by \(\frac{n}{n^2} = \frac{1}{n}\), which goes to 0 as \(n\) increases.

Thus, we have:
\[ \lim_{x \to \infty} \frac{xn(n + 1)}{2n^2} = \frac{(n + 1)x}{2n} \]

### Conclusion
As \(x\) goes to infinity, the limit approaches:
\[ \frac{(n + 1)}{2} \]

Therefore, the final answer is:
\[ \lim_{x \to \infty} \frac{[x] + [2x] + [3x] + \ldots + [nx]}{n^2} = \frac{n + 1}{2} \]